A short bar magnet has a magnetic moment m . At a far point on its eq…

The correct answer is Option (3) → antiparallel to the magnetic moment of the magnet

For a short bar magnet, the field at a far point on the normal bisector is \( \mathbf{B}_E=-\mu_0\mathbf{m}/4\pi r^{3} \). The minus sign shows the direction of the field.

Option (1) parallel to the magnetic moment of the magnet — Incorrect. The field points along \( \mathbf{m} \) on the axis of the magnet, where \( \mathbf{B}_A=\mu_0 2\mathbf{m}/4\pi r^{3} \). On the equatorial line it has the opposite sign.

Option (2) along the equatorial line, away from the magnet — Incorrect. \( \mathbf{B}_E \) is a multiple of \( \mathbf{m} \), so it is parallel to the axis of the magnet. It has no part along the equatorial line.

Option (3) antiparallel to the magnetic moment of the magnet — Correct. Because of the minus sign in \( \mathbf{B}_E \), the field points opposite to \( \mathbf{m} \), from the N-pole side towards the S-pole side.

Option (4) along the equatorial line, towards the magnet — Incorrect. A field pointing in to the magnet would be the field of a single pole at its centre. The two poles together give \( \mathbf{B}_E \) along \( -\mathbf{m} \), which is across the equatorial line.

NCERT (2026-27):
The electrostatic analog – “In particular, we can write down the equatorial field (B_E) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet:”

    \( \displaystyle \mathbf{B}_E=-\frac{\mu_0\mathbf{m}}{4\pi r^{3}} \)\( \qquad (5.4) \)